Difference between revisions of "User:Pifinity"
| (17 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
| − | == pifinity's problem set == | + | = pifinity's problem set = |
| + | <b>Welcome to pifinity's problem set! This page contains problems created by pifinity. Please do not edit this page.</b> | ||
| + | == Finalized Problems == | ||
| + | ==== Problem I ==== | ||
| + | Evaluate <math>\frac{48}{32} + \frac{24}{16}.</math> | ||
| − | + | ==== Solution I ==== | |
| + | Simplify; no denominator conversion needed: <cmath>\frac{48}{32} + \frac{24}{16}=\frac{3}{2} + \frac{3}{2} = \boxed{3}.</cmath> | ||
| − | <cite><sub>This page has | + | |
| + | == Unconfirmed Problems == | ||
| + | ==== Problem ==== | ||
| + | What is the <math>``\,?\,"</math> term? <cmath>(100-96)+100+(96-92)+94+(92-88)+88+\hdots+?+(4-0)</cmath> | ||
| + | ==== Solution ==== | ||
| + | We first wish to find the number of "telescoping" terms. The first one is <math>100</math> minus something, so we can label it <math>\tfrac{100}4=25.</math> The last term is <math>4</math> minus something, so we can label it <math>\tfrac{4}4=1.</math> Thus there are <math>25-(4-1)=22</math> terms. Therefore, there are 21 gaps between the terms in parentheses. Since there is one less of the terms not in parentheses, then there are 20 gaps between these terms in total. And since <math>6</math> is subtracted each time for the terms not in parentheses, then we have to subtract <math>6\times20=120</math> from <math>100</math>: <cmath>100-120=\boxed{-20}.</cmath> | ||
| + | == Unsolved Problems == | ||
| + | <b>No Unsolved Problems.</b> | ||
| + | |||
| + | = = | ||
| + | <cite><sub>This page has 12 official edits. The last edit was on 12:09, May 18, 2018.</sub></cite> | ||
Latest revision as of 14:09, 18 May 2018
Contents
pifinity's problem set
Welcome to pifinity's problem set! This page contains problems created by pifinity. Please do not edit this page.
Finalized Problems
Problem I
Evaluate 
Solution I
Simplify; no denominator conversion needed: ![\[\frac{48}{32} + \frac{24}{16}=\frac{3}{2} + \frac{3}{2} = \boxed{3}.\]](http://latex.artofproblemsolving.com/7/1/6/716ec936a2c62395a84fadfe52b137eab0761a06.png)
Unconfirmed Problems
Problem
What is the 
term? ![\[(100-96)+100+(96-92)+94+(92-88)+88+\hdots+?+(4-0)\]](http://latex.artofproblemsolving.com/f/8/8/f88425f55a1a400f2f6c44318dd12e4df4fcf617.png)
Solution
We first wish to find the number of "telescoping" terms. The first one is 
minus something, so we can label it 
The last term is 
minus something, so we can label it 
Thus there are 
terms. Therefore, there are 21 gaps between the terms in parentheses. Since there is one less of the terms not in parentheses, then there are 20 gaps between these terms in total. And since 
is subtracted each time for the terms not in parentheses, then we have to subtract 
from : ![\[100-120=\boxed{-20}.\]](http://latex.artofproblemsolving.com/7/a/4/7a4d8b725a685780fec1a079fa1690a3dfb213e4.png)
Unsolved Problems
No Unsolved Problems.
This page has 12 official edits. The last edit was on 12:09, May 18, 2018.
